JEE Main 2025. Previous Year Question.
Problem
Let $I = \displaystyle\int \dfrac{dx}{(x-1)^{11/13}(x+15)^{15/13}}$. Then $I$ is?
Key insight. The two fractional exponents $\tfrac{11}{13}$ and $\tfrac{15}{13}$ add up to exactly $2$ — so after substituting $t=x-1$ and factoring $t^{15/13}$ out of $(t+16)^{15/13}$, the leftover $t^{11/13}\cdot t^{15/13}$ collapses into a clean $t^2$. That’s the signal this substitution was designed to work.
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Approach
Substitute $t=x-1$, turning $x+15$ into $t+16$. Factor $t^{15/13}$ out of $(t+16)^{15/13}$, which combines with the existing $t^{11/13}$ in the denominator to give a clean $t^2$. Substitute again, this time for the bracketed expression $1+16/t$, reducing the integral to a simple power rule. Substitute back twice to express the answer in terms of $x$.
Solution
Step 1 — Substitute t = x − 1
$$I = \int \frac{dt}{t^{11/13}(t+16)^{15/13}}$$
Step 2 — Factor t^(15/13) out of (t + 16)^(15/13)
$$(t+16)^{15/13} = t^{15/13}\left(1+\frac{16}{t}\right)^{15/13}$$
$$I = \int \frac{dt}{t^{11/13}\cdot t^{15/13}\left(1+\dfrac{16}{t}\right)^{15/13}} = \int \frac{dt}{t^{26/13}\left(1+\dfrac{16}{t}\right)^{15/13}} = \int \frac{dt}{t^2\left(1+\dfrac{16}{t}\right)^{15/13}}$$
(using $\tfrac{11}{13}+\tfrac{15}{13}=\tfrac{26}{13}=2$)
Step 3 — Substitute z = 1 + 16/t
$$dz = -\frac{16}{t^2}\,dt \implies \frac{dt}{t^2} = -\frac{dz}{16}$$
$$I = \int \left(-\frac{1}{16}\right)z^{-15/13}\,dz = -\frac{1}{16}\int z^{-15/13}\,dz$$
Step 4 — Integrate using the power rule
$$-\frac{1}{16}\cdot\frac{z^{-15/13+1}}{-15/13+1} + C = -\frac{1}{16}\cdot\frac{z^{-2/13}}{-2/13}+C = \frac{13}{32}z^{-2/13}+C$$
Step 5 — Substitute back z = 1 + 16/t = (t+16)/t
$$z^{-2/13} = \left(\frac{t+16}{t}\right)^{-2/13} = \left(\frac{t}{t+16}\right)^{2/13} = \frac{t^{2/13}}{(t+16)^{2/13}}$$
$$I = \frac{13}{32}\cdot\frac{t^{2/13}}{(t+16)^{2/13}}+C$$
Step 6 — Substitute back t = x − 1 (so t + 16 = x + 15)
$$I = \frac{13}{32}\cdot\frac{(x-1)^{2/13}}{(x+15)^{2/13}}+C$$
Answer
$$I = \frac{13}{32}\cdot\frac{(x-1)^{2/13}}{(x+15)^{2/13}}+C$$
Common mistakes
- Missing that the two exponents add to exactly 2. This isn’t a coincidence — it’s the entire reason the substitution $t=x-1$ (rather than any other shift) turns the denominator into a clean $t^2$ times a bracket. Spotting this relationship between $\tfrac{11}{13}$ and $\tfrac{15}{13}$ upfront is what makes the rest of the problem tractable.
- Forgetting the negative sign when substituting $dt/t^2=-dz/16$. Differentiating $z=1+16/t$ gives $dz=-16/t^2\,dt$, so $dt/t^2=-dz/16$ — dropping this sign flips the sign of the final answer.
Practise next
- Solve $\displaystyle\int \dfrac{dx}{(x-2)^{9/11}(x+9)^{13/11}}$, using the same shift-factor-substitute approach (checking first that the two exponents add to $2$).
Show answer
$\dfrac12\left(\dfrac{x-2}{x+9}\right)^{2/11}+C$. The exponents add to $\tfrac{9}{11}+\tfrac{13}{11}=2$, which is the signal for this method.
Factor out $(x+9)^2$: the integrand is $\dfrac{1}{(x+9)^2}\left(\dfrac{x-2}{x+9}\right)^{-9/11}$.
With $u=\dfrac{x-2}{x+9}$, $du=\dfrac{11}{(x+9)^2}dx$, so the integral is $\dfrac1{11}\displaystyle\int u^{-9/11}du=\dfrac1{11}\cdot\dfrac{11}{2}u^{2/11}+C$. Differentiating the result gives back the integrand exactly.
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